LeetCode 1760. Minimum Limit of Balls in a Bag

Binary Search

You are given an integer array nums where the ith bag contains nums[i] balls. You are also given an integer maxOperations.

You can perform the following operation at most maxOperations times:

• Take any bag of balls and divide it into two new bags with a positive number of balls.

  • For example, a bag of 5 balls can become two new bags of 1 and 4 balls, or two new bags of 2 and 3 balls.

Your penalty is the maximum number of balls in a bag. You want to minimize your penalty after the operations.

Return the minimum possible penalty after performing the operations.

Example 1:

Input: nums = [9], maxOperations = 2
Output: 3
Explanation: 
- Divide the bag with 9 balls into two bags of sizes 6 and 3. [9] -> [6,3].
- Divide the bag with 6 balls into two bags of sizes 3 and 3. [6,3] -> [3,3,3].
The bag with the most number of balls has 3 balls, so your penalty is 3 and you should return 3.

Example 2:

Input: nums = [2,4,8,2], maxOperations = 4
Output: 2
Explanation:
- Divide the bag with 8 balls into two bags of sizes 4 and 4. [2,4,8,2] -> [2,4,4,4,2].
- Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,4,4,4,2] -> [2,2,2,4,4,2].
- Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,2,2,4,4,2] -> [2,2,2,2,2,4,2].
- Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,2,2,2,2,4,2] -> [2,2,2,2,2,2,2,2].
The bag with the most number of balls has 2 balls, so your penalty is 2 an you should return 2.

Example 3:

Input: nums = [7,17], maxOperations = 2
Output: 7

Constraints:

• 1 <= nums.length <= 10^5

• 1 <= maxOperations, nums[i] <= 10^9

Solution:

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class Solution {
    
    bool check(const vector<int>& nums, int maxOperations, int penalty) {
        int ops = 0;
        for (int num : nums) {
            ops += (num-1) / penalty;
        }
        return ops <= maxOperations;
    }
    
public:
    int minimumSize(vector<int>& nums, int maxOperations) {
        int left = 1, right = *max_element(nums.begin(), nums.end());
        
        while (left < right) {
            int mid = (left + right) / 2;
            
            bool can_do_it = check(nums, maxOperations, mid);
            
            if (can_do_it) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        
        return left;
    }
};